对元组列表中的每个值求和 [英] sum each value in a list of tuples
本文介绍了对元组列表中的每个值求和的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个与此类似的元组列表:
I have a list of tuples similar to this:
l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
我想创建一个简单的单行,它会给我以下结果:
I want to create a simple one-liner that will give me the following result:
r = (25, 20) or r = [25, 20] # don't care if tuple or list.
这会像执行以下操作:
r = [0, 0]
for t in l:
r[0]+=t[0]
r[1]+=t[1]
我相信这是很简单的事情,但我想不出来.
I am sure it is something very simple, but I can't think of it.
注意:我已经看过类似的问题了:
Note: I looked at similar questions already:
如何对 Python 中元组列表中的每个元组的第一个值求和?
推荐答案
使用zip()
和sum()
:
In [1]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]
In [2]: [sum(x) for x in zip(*l)]
Out[2]: [25, 20]
或:
In [4]: map(sum, zip(*l))
Out[4]: [25, 20]
timeit
结果:
In [16]: l = [(1, 2), (3, 4), (5, 6), (7, 8), (9, 0)]*1000
In [17]: %timeit [sum(x) for x in zip(*l)]
1000 loops, best of 3: 1.46 ms per loop
In [18]: %timeit [sum(x) for x in izip(*l)] #prefer itertools.izip
1000 loops, best of 3: 1.28 ms per loop
In [19]: %timeit map(sum, zip(*l))
100 loops, best of 3: 1.48 ms per loop
In [20]: %timeit map(sum, izip(*l)) #prefer itertools.izip
1000 loops, best of 3: 1.29 ms per loop
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